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**Hint:**Here, we will use the step-deviation method of finding the Mean. We will find the Class marks or the midpoint and then the deviation from midpoint, sum of frequencies, and the sum of the product of frequencies and their respective class marks. Substituting these in the formula of Mean, we will find the missing value of the frequency $f$. Mean is actually the average of the given numbers.

**Complete step-by-step answer:**

According to the question,

We are given the mean profit per shop and we are required to find the missing frequency $f$.

We will use the step-deviation method to find the mean of this question.

According to this method, we will find the midpoint of the given class intervals.

Also, let $a = 25$and here, width of class, $h = 10$

Mid-point or \[m\] is equal to the sum of lower limit and upper limit divided by 2.

Hence, we will draw the following table:

Profit per shops | Number of shops${f_i}$ | Mid-Point${x_i}$ | Deviation${u_i} = \dfrac{{{x_i} - 25}}{{10}}$ | ${f_i}{u_i}$ |

0-10 | 12 | $\dfrac{{0 + 10}}{2} = \dfrac{{10}}{2} = 5$ | $\dfrac{{5 - 25}}{{10}} = \dfrac{{ - 20}}{{10}} = - 2$ | $12 \times - 2 = - 24$ |

10-20 | 18 | $\dfrac{{10 + 20}}{2} = \dfrac{{30}}{2} = 15$ | $\dfrac{{15 - 25}}{{10}} = \dfrac{{ - 10}}{{10}} = - 1$ | $18 \times - 1 = - 18$ |

20-30 | 27 | $\dfrac{{20 + 30}}{2} = \dfrac{{50}}{2} = 25$ | $\dfrac{{25 - 25}}{{10}} = 0$ | $27 \times 0 = 0$ |

30-40 | f | $\dfrac{{30 + 40}}{2} = \dfrac{{70}}{2} = 35$ | $\dfrac{{35 - 25}}{{10}} = \dfrac{{10}}{{10}} = 1$ | $f \times 1 = f$ |

40-50 | 17 | $\dfrac{{40 + 50}}{2} = \dfrac{{90}}{2} = 45$ | $\dfrac{{45 - 25}}{{10}} = \dfrac{{20}}{{10}} = 2$ | $17 \times 2 = 34$ |

50-60 | 6 | $\dfrac{{50 + 60}}{2} = \dfrac{{110}}{2} = 55$ | $\dfrac{{55 - 25}}{{10}} = \dfrac{{30}}{{10}} = 3$ | $6 \times 3 = 18$ |

Total: | $\sum {{f_i} = } 80 + f$ | $\sum {{f_i}{u_i} = } 10 + f$ |

Also, we know that the formula of Mean by step-deviation method $ = a + \dfrac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }} \times h$

Hence, we have calculated the total i.e. $\sum {{f_i} = } 80 + f$ and $\sum {{f_i}{u_i} = } 10 + f$

Mean $ = 25 + \dfrac{{10 + f}}{{80 + f}} \times 10$…………………………$\left( 1 \right)$

But according to the question,

Mean $ = 28$

Substituting this in equation $\left( 1 \right)$, we get

$ \Rightarrow 28 = 25 + \dfrac{{10 + f}}{{80 + f}} \times 10$

Subtracting 25 from both sides, we get

$ \Rightarrow 3 = \dfrac{{100 + 10f}}{{80 + f}}$

Taking the denominator from RHS to LHS, we get

$ \Rightarrow 3\left( {80 + f} \right) = 100 + 10f$

$ \Rightarrow 240 + 3f = 100 + 10f$

Solving further, we get

$ \Rightarrow 7f = 140$

Dividing both sides by 7, we get

$ \Rightarrow f = 20$

**Hence, the required value of missing frequency $f$ is 20.**

**Note:**

It is usually stated as the sum of the observations divided by the total number of observations. In this question as well, we have multiplied the frequencies by the class marks which actually represent the marks of the whole class falling in that particular frequency. By adding all the products of the frequencies and the class marks, we actually find the ‘sum of observations’. And finally, when we divide this by the sum of frequencies, we get our required mean as the sum of frequencies gives us the total number of observations. Hence, no matter which formula is being used in which question, the root technique and meaning remains the same.

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